First slide

Methods of integration

Question

$E v a l u a t e \int 2^{2^{2^{x}}} 2^{2^{x}} 2^{x} d x$

Easy

A

$= \frac{1}{(\log 2)^{3}} 2^{2^{x}} + c$
B

$= \frac{1}{(\log 2)^{3}} 2^{2^{2^{x}}} + c$
C

$= \frac{1}{(\log 2)^{3}} 2^{x} + c$
D

$= \frac{1}{(\log 2)^{2}} 2^{2^{2^{x}}} + c$

Solution

$\begin{array}{l} I = \int 2^{2^{2^{x}}} 2^{2^{x}} 2^{x} d x \\ Let 2^{2^{2^{x}}} = t or 2^{2^{2^{x}}} 2^{2^{x}} 2^{x} (\log 2)^{3} d x = d t \\ \begin{aligned} ∴ I = \int \frac{1}{(\log 2)^{3}} d t & = \frac{1}{(\log 2)^{3}} t + c \end{aligned} \\ = \frac{1}{(\log 2)^{3}} 2^{2^{2^{x}}} + c \end{array}$

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